Question

Let a, b, c, d, e be the observations with mean m and standard deviation s. The standard deviation of the observations a + k, b + k, c + k, d + k, e + k is

विकल्प

• s     

• ks    

•  s + k    

• \[\frac{s}{k}\]

Answer 1

The given observations are a, b, c, d, e.
Mean = m =\[\frac{a + b + c + d + e}{5}\]

\[\Rightarrow \sum_{} x_i = a + b + c + d + e = 5m\]      .....(1)

Standard deviation, s = \[\sqrt{\frac{\sum_{} x_i^2}{5} - m^2}\]

Now, consider the observations a + k, b + k, c + k, d + k, e + k.
New mean

\[= \frac{\left( a + k \right) + \left( b + k \right) + \left( c + k \right) + \left( d + k \right) + \left( e + k \right)}{5}\]

\[= \frac{a + b + c + d + e + 5k}{5}\]

\[ = \frac{5m + 5k}{5}\]

\[ = m + k\]

∴ New standard deviation

\[= \sqrt{\frac{\sum_{} \left( x_i + k \right)^2}{5} - \left( m + k \right)^2}\]

\[ = \sqrt{\frac{\sum_{} \left( x_i^2 + k^2 + 2 x_i k \right)}{5} - \left( m^2 + k^2 + 2mk \right)}\]

\[ = \sqrt{\frac{\sum_{} x_i^2}{5} + \frac{\sum_{} k^2}{5} + \frac{\sum_{} 2 x_i k}{5} - \left( m^2 + k^2 + 2mk \right)}\]

\[ = \sqrt{\frac{\sum_{} x_i^2}{5} - m^2 + \frac{5 k^2}{5} - k^2 + \frac{2k \sum_{} x_i}{5} - 2mk}\]

\[= \sqrt{\frac{\sum_{} x_i^2}{5} - m^2 + \frac{2k \times 5m}{5} - 2mk} \left[ \text{ Using } \left( 1 \right) \right]\]

\[ = \sqrt{\frac{\sum_{} x_i^2}{5} - m^2}\]

\[ = s\]